The product of n consecutive integers is divisible by n factorial

Question

How can we prove that the product of n consecutive integers is divisible by n factorial?

Note: In this subsequent question and the comments here the OP has clarified that he seeks a proof that "does not use the properties of binomial coefficients". Please post answers in said newer thread so that this incorrectly-posed question may be closed as a duplicate.

Answer 1

This is almost immediate from the fact that the binomial coefficient $$\binom{k+n}{k}$$ is an integer. Just write the product $(k+1) \cdots (k+n)$ accordingly and you'll have your answer.

Answer 2

You might be interested in this blog post of Timothy Gowers:

http://gowers.wordpress.com/2010/09/18/are-these-the-same-proof/

Answer 3

Let us prove that $m^{(k)}=m(m+1)...(m+k-1)$ is divided by $k!$ for all integer $m$. Induction by $k$.

$k=1$: Every integer $m$ is divided by $1$

$k\to k+1$:

• induction by $m$: $m=0$: $0^{k+1}=0$ is divided by $(k+1)!$

  $m\to m+1$: $(m+1)^{(k+1)}=((m+1)^{(k+1)}-m^{(k+1)})+m^{(k+1)}=$

  $=(m+1)(m+2)...(m+k+1)-m(m+1)...(m+k)+m^{(k+1)}=$

  $(k+1)(m+1)...(m+k)+m^{(k+1)}=(k+1)(m+1)^{(k)}+m^{(k+1)}$

  and first term is divided by $(k+1)\cdot k!=(k+1)!$ because of induction by k and the second term is divided by $(k+1)!$ because of induction by $m$

  the same works for $m\to m-1$

Update Oops, essentially the same proof found in the thread mentioned in this answer.

Answer 4

A clearer version of NurdinTakenov's proof. I prefer Knuth's notation, and falling factorials are nicer to work with: $$ m^{\underline{k}} = m (m - 1) \ldots (m - k + 1) $$

First: $$ (m + 1)^{\underline{k}} - m^{\underline{k}} = (m + 1) \cdot m^{\underline{k - 1}} - m^{\underline{k - 1}} \cdot (m - k + 1) \\ = k \cdot m^{\underline{k - 1}} $$ So: $$ \sum_{1 \le r \le m} r^{\underline{k}} = \frac{m^{\underline{k + 1}}}{k + 1} $$ Now the proof by induction over $k$ goes through easily:

Base: If $k = 0$, we have that $0! \mid m^{\underline{0}}$, which is just $1 \mid 1$.

Induction: Assume $k! \mid m^{\underline{k}}$ for all $m$. Then: $$ m^{\underline{k + 1}} = (k + 1) \sum_{1 \le r \le m} r^{\underline{k}} $$ By induction, each term of the sum is divisible by $k!$, so the right hand side is divisible by $(k + 1) k! = (k + 1)!$.

Answer 5

The identity below shows that the problem is equivalent to the fact that binomial coefficients are integral - for which various proofs are known, e.g. using their recursion, or their well-known combinatorial interpretation, or their minimality in terms of prime divisors - see this prior question

$$\rm\displaystyle\quad\quad {m \choose n}\ =\ \frac{m!/(m-n)!}{n!}\ =\ \frac{m\:(m-1)\:\cdots\:(m-n+1)}{n\:(n-1)\quad\quad\:\cdots\:\quad\quad 1\quad\quad}$$